Isn't this just the definition of a uniformly continuous function?
I thought just plain old continuous was the local definition, and uniformly continuous was the global definition.I think the definition of uniformly continuous is a local definition
Not really; it's a property that must be satisfied at all values of ##x##, but so is ordinary continuity. But the definition of the property, for any given value of ##x##, only involves that value of ##x##; it does not involve the whole set of values of ##x## all at once.The problem statement here is a statement about the whole function
I don't think so, because a uniformly continuous function does not have to be linear. The problem seems to be asking you to show that any uniformly continuous function can be approximated by a linear function.Isn't this just the definition of a uniformly continuous function?
Once you pick a choice of ##\epsilon## uniform continuity only says things about points that are close together. The fact that the same Delta and epsilon work is global, but the definition still says if two inputs are close together, so are the outputs.I thought just plain old continuous was the local definition, and uniformly continuous was the global definition.
Not really; it's a property that must be satisfied at all values of ##x##, but so is ordinary continuity. But the definition of the property, for any given value of ##x##, only involves that value of ##x##; it does not involve the whole set of values of ##x## all at once.
This is just as true of the problem statement in the OP. So I don't see that there is a useful "local" vs. "global" distinction here.Once you pick a choice of ##\epsilon## uniform continuity only says things about points that are close together. The fact that the same Delta and epsilon work is global, but the definition still says if two inputs are close together, so are the outputs.
The OP statement doesn't compare the function's values at two different values of ##x## at all.the Op immediately gives a way to compare the output of two inputs no matter how far away they are
I'm not sure this proof is valid.@anuttarasammyak's proof
It's conceptually a bit messy, IMO, as it leaves ##\epsilon## arbitrary. He should have let ##\epsilon =1##, for example.I'm not sure this proof is valid.
What do you all think of ChatGPT's answer below?Let ##f : \mathbb{R} \to \mathbb{R}## be a uniformly continuous function. Show that, for some positive constants ##A## and ##B##, we have ##|f(x)| \le A + B|x|## for all ##x\in \mathbb{R}##.
Since ##f## is uniformly continuous, there exists a constant ##\delta > 0## such that, for any ##x,y \in \mathbb{R}## with ##|x - y| < \delta##, we have
$$|f(x) - f(y)| < 1.$$
Take any ##x \in \mathbb{R}##. Then, for any ##y \in \mathbb{R}## with ##|y| \le |x| + \delta##, we have
$$|f(x) - f(y)| < 1.$$
In particular, if we take ##y = 0##, we get
$$|f(x)| \le 1 + |f(0)|.$$
Therefore, we can take ##A = |f(0)|## and ##B = 1##. This shows that, for some positive constants ##A## and ##B##, we have
$$|f(x)| \le A + B|x|$$
for all ##x \in \mathbb{R}##.
Note that the choice of ##A## and ##B## is not unique: any constants ##A'## and ##B'## such that ##A' \ge A## and ##B' \ge B## will also work. For example, we can take ##A' = \max {|f(x)| : x \in \mathbb{R}}## and ##B' = \delta^{-1}##, where ##\delta## is the constant from the uniform continuity of ##f##.
Bounded derivative is sufficient, per the mvt ( though function does not have to be differentiable to be UC.I don't think so, because a uniformly continuous function does not have to be linear. The problem seems to be asking you to show that any uniformly continuous function can be approximated by a linear function.